plate properties (PLAXIS 2d vs PLAXIS 3D)

Hi Faisal,

In PLAXIS 3D, all necessary parameters (dimensions and unit weight) are input parameters: we define a plate material with a specific height (d) and unit weight (γ), and we assign that material to a polygon with known width and length (b x L). 

In PLAXIS 2D, the specific weight is the force per unit length and per unit width in the out-of-plane direction (expressed in kN/m/m units). It represents the product of the material’s unit weight (γ) over the area of the cross-section: w = γ × b × d (kN/m). In a plane strain analysis, we consider 1m in the out-of-plane direction; therefore, b = 1m, and so w = γ × d (kN/m/m). 

This is also what the manual explains by saying that the specific weight is obtained by multiplying the unit weight of the plate material (γ) by the thickness of the plate (d). Therefore, the first “/m” corresponds to the length (distributed load), and the second “/m” is assumed to be the out-of-plane width of 1m.

Let us take the following example: suppose you have a steel plate (b x d x L) = (b x 1.5m x 40m). Let us assume that the density of steel is approximately ρ = 7850 kg/m3 ≈ 78.5 kN/m3. Therefore, the total weight of the plate is: W = ρ * V = ρ * (b * d * L) = 78.5 * 1.5 * 40 = 4710 (kN/m). Here the “/m” implies that this is the total weight of the plate is for 1m in the out-of-plane direction. If we divide W by its length, we get the unit weight (load per unit length per unit width in the out-of-plane direction), which is essentially the unit weight in PLAXIS 2D: w = W/L = 117.75 (kN/m/m).

Hello Vasileios

According to the Plaxis 2D V21 manual, the weight of the soil should also be subtracted from the weight of the plate, in order to get correct unit weight w. Is this correct:

For relatively massive structures the weight of a plate is, in principle, obtained by multiplying the unit weight of
the plate material by the thickness of the plate. Note that in a finite element model, plates are superimposed on a
continuum and therefore 'overlap' the soil. To calculate accurately the total weight of soil and structures in the
model, the unit weight of the soil should be subtracted from the unit weight of the plate material. For sheet-pile
walls the weight (force per unit area) is generally provided by the manufacturer. This value can be adopted
directly since sheet-pile walls usually occupy relatively little volume.

Hi Sasa,
Indeed, the weight of the plate element is not the full weight of the structures, but it is the weight of the structures minus out the weight of the soil removed. See also: https://communities.bentley.com/products/geotech-analysis/w/wiki/58294/considerations-for-modelling-a-thick-slab-as-a-plate

Thank you for clarification.

What about the case, where one side of the plate element is excavated, does the model take that into account?

Or the opposite case, when the unit weight of the rock (26kN/m3) is larger than the unit weight of the concrete in plate element (24kN/m3)?

Thank you Engineer Vasileios Basas  for your clarification, 

Will the specific weight differ if the plate you mentioned (b x 1.5 x 40) was rotated 90 degress, as if the plate now is drawn vertically as a diaphram wall (instead of a footing), will the specifc weight in that case be w = 4710/1.5 = 3140 (KN/m/m), or will it still be equal = 117.5 (KN/m/m), 

Please advice on that matter, Thank you

In case of tunnel modelling, unit weight of soil should be subtracted from the unit weight of concrete?

In case of tunnel modelling, unit weight of soil should be subtracted from the unit weight of concrete?

You just need to think what you are aiming. The link Vasileios is really good. For some cases where the bending moments are more important, I use concrete unit weight x plate thickness for 2D. However, this creates higher pressure in the soil, so settlements are higher. If settlements are important, you go for correct stress in soil, so you need to reduce the soil unit weight. 

Not an easy task always.

ok sure,thanks